Resolving Rational Fractions into Partial Fractions

Introduction

Resolving into partial fractions means expressing a complicated rational fraction as the sum of simpler fractions. This makes algebraic expressions easier to work with, particularly when solving integration problems.

For example:

5 / [x(x + 2)] = 5/(2x) − 5/[2(x + 2)]

To resolve a rational fraction correctly, first check whether it is proper, factorize its denominator, select the correct partial fraction form, and then calculate the unknown constants.

1. Steps for Resolving Partial Fractions

Follow these steps whenever you solve a partial fraction problem.

1. Check the fraction: If the degree of the numerator is greater than or equal to the degree of the denominator, perform polynomial division first.

2. Factorize the denominator: Express it as a product of linear and/or irreducible quadratic factors.

3. Write the assumed decomposition: Select the correct form based on the denominator factors.

4. Multiply by the common denominator: This removes the fractions and produces a polynomial equation.

5. Calculate the unknown constants: Substitute suitable values of x or compare coefficients.

6. Write the final answer: Substitute the constants into the assumed decomposition.

2. Solved Examples

Example 1 — Basic Level

Question: Resolve the following into partial fractions.

3 / [x(x + 1)]

Step 1: Write the assumed form.

3 / [x(x + 1)] = A/x + B/(x + 1)

Step 2: Multiply by x(x + 1).

3 = A(x + 1) + Bx

Expanding:

3 = (A + B)x + A

Step 3: Compare coefficients.

Comparing constant terms:

A = 3

Comparing the coefficients of x:

A + B = 0

Substitute A = 3:

3 + B = 0

B = −3

Final answer:

3 / [x(x + 1)] = 3/x − 3/(x + 1)

Example 2 — Medium Level

Question: Resolve the following into partial fractions.

(2x + 7) / (x + 2)²

Step 1: Write the assumed form.

Because the factor (x + 2) is repeated twice:

(2x + 7) / (x + 2)² = A/(x + 2) + B/(x + 2)²

Step 2: Multiply by (x + 2)².

2x + 7 = A(x + 2) + B

Expanding:

2x + 7 = Ax + 2A + B

Step 3: Compare coefficients.

Coefficient of x:

A = 2

Constant terms:

2A + B = 7

Substitute A = 2:

4 + B = 7

B = 3

Final answer:

(2x + 7) / (x + 2)² = 2/(x + 2) + 3/(x + 2)²

Example 3 — Challenging Level

Question: Resolve the following into partial fractions.

(x² + 2x + 3) / [(x + 1)(x² + 1)]

Step 1: Write the assumed form.

The denominator contains a linear factor and an irreducible quadratic factor.

Therefore:

(x² + 2x + 3) / [(x + 1)(x² + 1)] = A/(x + 1) + (Bx + C)/(x² + 1)

Step 2: Multiply by the common denominator.

x² + 2x + 3 = A(x² + 1) + (Bx + C)(x + 1)

Expanding:

x² + 2x + 3 = Ax² + A + Bx² + Bx + Cx + C

Collecting terms:

x² + 2x + 3 = (A + B)x² + (B + C)x + (A + C)

Step 3: Compare coefficients.

A + B = 1

B + C = 2

A + C = 3

From the first equation:

B = 1 − A

From the second equation:

C = 2 − B = 1 + A

Substitute into the third equation:

A + (1 + A) = 3

2A = 2

A = 1

Therefore:

B = 1 − 1 = 0

C = 1 + 1 = 2

Final answer:

(x² + 2x + 3) / [(x + 1)(x² + 1)] = 1/(x + 1) + 2/(x² + 1)

3. Important Tips

4. Practice Questions

Try resolving these fractions independently.

Basic:

4 / [x(x + 3)]

Medium:

(5x + 4) / (x + 1)²

Challenging:

(2x² + 5x + 3) / [(x + 1)(x² + 1)]

Answers for Checking

• Basic: 4/(3x) − 4/[3(x + 3)]

• Medium: 5/(x + 1) − 1/(x + 1)²

• Challenging: 2/(x + 1) + 3x/(x² + 1)

Key Takeaways

Resolving into partial fractions involves factorizing the denominator, choosing the correct decomposition form, and finding the unknown constants through substitution or coefficient comparison. Regular practice with distinct, repeated and irreducible quadratic factors builds the foundation required for integration and other applications in Remedial Mathematics.